# Volume under sphere with radius 1, center (0, 0, 1) and above the cone $z = \sqrt{x^2+y^2}$

so I want to find the volume of the body D defined as the region under a sphere with radius 1 with the center in (0, 0, 1) and above the cone given by $z = \sqrt{x^2+y^2}$. The answer should be $\pi$. A hint is included that you should use spherical coordinates. I've started by making a equation for the sphere, $x^2+y^2+(z-1)^2=1$. I used the transformation $(x, y, z) = (\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi+1)$. Now I'm struggling on defining the region D. I got that $0\leq\theta\leq2\pi$, but I can't find the bounds for $\phi$ and $\rho$.

• Can you use cylindrical coordinates? I think that it's better. – Emilio Novati Jan 23 '17 at 17:37
• I would advice you to draw this region, then you'll see it clearly – Ismasou Jan 23 '17 at 17:47

## 2 Answers

If you try spherical coordinates, the following happens. For the cone, you have $$\rho\cos\phi=z=\sqrt{x^2+y^2}=\sqrt{\rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta}=\rho\sin\phi.$$(this assume that you chose $\phi$ so that $\sin\phi\geq0$, that is $0\leq\phi\leq\pi$). So $\cos\phi=\sin\phi$; it follows that the equation of cone is $\phi=\pi/4$.

So, to describe the interior of your region, you will have $0\leq\phi\leq\pi/4$; simple enough. From the equation of the sphere we get $$\rho^2=2\rho\cos\phi,$$ or $\rho=2\cos\phi$.

The volume is then \begin{align} V&=\iiint_E1\,dV=\int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\cos\phi}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ \ \\ &=\frac{16\pi}3\,\int_0^{\pi/4}\cos^3\phi\,\sin\phi\,d\phi =-\frac{16\pi}3\,\left.\left(\frac{\cos^4\phi}4 \right)\right|_0^{\pi/4}\\ \ \\ &=\frac{4\pi}3\left(1-\frac1{4} \right)=\pi. \end{align}

On the other hand, to work in cylindrical coordinates we have as follows. If we write $s=\sqrt{x^2+y^2}$, the intersection of the sphere and the cone happens when $s^2+(s-1)^2=1$, with solutions $s=0$ and $s=1$. As we want to be above the cone, we have to choose $s=1$. The equations of the sphere and the cone are, respectively, $z=1+\sqrt{1-r^2}$ and $z=r$. Then the volume is \begin{align} V&=\int_0^{2\pi}\int_0^1\,(1+\sqrt{1-r^2}-r)\,r\,dr\,d\theta =2\pi\,\int_0^1(r+r\sqrt{1-r^2}-r^2)\,dr\\ \ \\ &=2\pi\,\left.\left(\frac{r^2}2-\frac{\sqrt{1-r^2}}3-\frac{r^3}3\right)\right|_0^1 =2\pi\,\left(\frac12-0-\frac13+\frac13 \right)=\pi. \end{align}

• The equation of the sphere is $\rho=2\cos\phi$ if I am not mistaken. – Kuifje Jan 23 '17 at 18:04
• I think you forgot to simplify the $1's$ when expanding $(z-1)^2$. – Kuifje Jan 23 '17 at 18:06
• Yes, you are right. Thanks! – Martin Argerami Jan 23 '17 at 18:11

In spherical coordinates: $$E=\{(\rho,\theta,\phi)|0\le \theta\le 2\pi, 0\le \phi \le \pi/4, 0\le \rho \le 2\cos \phi \}$$ It follows that

$$\boxed{ V= \iiint_E dV = \int_0^{2\pi}\int_0^{\pi/4}\int_0^{2\cos\phi}\rho^2\sin\phi \;d\rho d\phi d\theta = \pi }$$

Note: to find your bounds for $\phi$ and $\rho$:

• The upper bound for $\phi$ is precisely the cone $z=\sqrt{x^2+y^2}\; \Leftrightarrow \; \rho \cos \phi = \rho \sin \phi \; \Leftrightarrow \; \phi = \pi/4$
• The upper bound for $\rho$ is precisely the sphere $x^2+y^2+(z-1)^2=1\; \Leftrightarrow \;\rho^2\sin^2\phi+(\rho \cos\phi-1)^2=1\; \Leftrightarrow \;\rho=2\cos\phi$
• Why does $\rho$ start at $0$ though? Doesen't that mean you get the volume of the part of the sphere thats underneath the cone too? Why shouldn't it go from the intersection of the cone and the sphere to the spherical shell? – Pame May 10 '18 at 10:18